SMALL INCREMENTS CALCULUS 13. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Small increments 1/15 Adrian Jannetta

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1 SMALL INCREMENTS CALCULUS 13 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Small increments 1/15 Adrian Jannetta

2 Objectives In this presentation we re going to look at one of the applications of partial differentiation. Absolute changes to a function. Relative changes to a function (percentage errors) How do these influence our predictions or measurements? (sensitivity analysis) By this stage you should be able to comfortably find all first partial derivatives for a given function. Small increments 2/15 Adrian Jannetta

3 Small changes Consider a rectangle with a length and width of x and y. The area A of the rectangle is A=xy x y If we increase the dimensions by small amountsδx andδy then we ll also increase the area by a small amountδa. δy y x δx The area of this new rectangle is related to the original one by A+δA=(x+δx)(y+δy) Small increments 3/15 Adrian Jannetta

4 We re going to investigate the relationship betweenδa and the changesδx and δy. A+δA=(x+δx)(y+δy) Multiply the brackets out on the RHS: A+δA=xy+ yδx+xδy+δxδy Each of the terms on the RHS corresponds to one of the smaller regions in the picture below. δy y x δx Small increments 4/15 Adrian Jannetta

5 Now since A=xy then those terms cancel out: δa=yδx+xδy+δxδy If the changesδx andδy are small then the productδxδy is the smallest term on the RHS. To a good approximation we can say: δa yδx+xδy (1) Take a look at the formula for area we started with and notice that the two partial derivatives are A x = y and A y = x Therefore we express equation (1) in these terms: δa A x δx+ A δy (2) y This gives the change in area as a relationship between the partial derivatives and the small changes themselves. Small increments 5/15 Adrian Jannetta

6 Small changes (again) Let s choose another shape and try again. Consider the case of a cylinder with radius r and height h. h The volume V is given by V=πr 2 h r We can write down partial derivatives of the volume: V h =πr2 and V r = 2πrh Small increments 6/15 Adrian Jannetta

7 If we make small changes to the radius (δr) and height (δh), then the volume will also change by a small amountδv Now expand the brackets: V+δV=π(r+δr) 2 (h+δh) V+δV V+δV = π(r 2 + 2rδr+(δr) 2 )(h+δh) = (πr 2 + 2πrδr+π(δr) 2 )(h+δh) = πr 2 h+2πrhδr+πh(δr) 2 +πr 2 δh+2πrδrδh+π(δr) 2 δh We know V=πr 2 h so it will cancel from both sides: δv= 2πrhδr+πh(δr) 2 +πr 2 δh+2πrδrδh+π(δr) 2 δh Theδ quanties are small, which means terms like(δr) 2 orδrδh are even smaller. We re going to ignore them now but doing so means the relationship is no longer exact: δv 2πrhδr+πr 2 δh Small increments 7/15 Adrian Jannetta

8 The volume of a cylinder is V=πr 2 h. The partial derivatives are: V r = 2πrh and V h =πr2 Let s examine that formula forδv for the cylinder: δv 2πrhδr+πr 2 δh It contains the partial derivatives of V : δv V r δr+ V h δh This means that the small change in volume is related to the small changes in the radius and height and we can can calculate the overall change using the partial derivatives; we don t have to expand the brackets in the way we did here! Small increments 8/15 Adrian Jannetta

9 Chain rule for small changes It doesn t matter whether we start with a rectangle, cylinder or some other formula. For a function of two variables u=u(x, y) small changes in x and y will produce a change in the overall value of u given by δu u x δx+ u y δy For a function of three variables u=u(x, y, z) we add another link to the chain: δu u x δx+ u y δy+ u z δz For functions of four or more variables the pattern continues in the same way. Small increments 9/15 Adrian Jannetta

10 Small changes Consider a cylinder of radius 3 cm and height 2 cm. How does the volume change when the radius is increased by 0.1 cm and the height is decreased by 0.2 cm? We saw that the change in volume is given by δv 2πrhδr+πr 2 δh We substitute the radius and height r= 3, h=2 into this along with the small changesδr= 0.1 andδh= 0.2. δv 2π(3)(2)(0.1)+π(3 2 )( 0.2) 1.2π 1.8π 0.6π The volume decreases by approximately 0.6π (about 1.88cm 3 ). Small increments 10/15 Adrian Jannetta

11 A right angle triangle A right angled triangle has shorter sides of length a = 3 cm and b = 4 cm. Use the chain rule to calculate the change to the hypotenuse if a is increased by 0.1 cm and b is decreased by 0.2 cm. The hypotenuse length is given by c=(a 2 + b 2 ) 1 2. The partial derivatives with respect to the two variables are: c a = a a 2+ b, c 2 b = b a 2+ b 2 It is given that δa = 0.1 and δb = 0.2. The chain rule for this problem is: Substitution gives: δc δc c a δa+ c b δb a b a 2+ b 2δa+ a 2+ b 2δb (0.1) ( 0.2) (0.6)(0.1)+(0.8)( 0.2) δc 0.1 The small changes give an approximate 0.1 cm decrease in the hypotenuse. Small increments 11/15 Adrian Jannetta

12 Percentage errors A force F between two masses M and m separated by a distance r is given by F= GMm r 2 where G is a constant. Suppose errors of±1% are possible in measurements of M, m and r, find the maximum possible error in the calculated value of F. The chain rule for this situation is δf F M δm+ F m δm+ F r δr The measurement errors are δm = ±0.01M, δm = ±0.01m and δr = ±0.01r. Substituting these, with the partial derivatives, into the formula gives: δf Gm r 2 (±0.01M)+ GM r 2 The variables can be simplified in each term: δf GMm r 2 (±0.01m)+ 2GMm (±0.01r) r 3 (±0.01)+ GMm (±0.01)+ 2GMm (±0.01) r 2 r 2 Small increments 12/15 Adrian Jannetta

13 In each term we have the original expression for F: δf F(±0.01) + F(±0.01) 2F(±0.01) F(±0.01± ) To get maximum possible error combine the±terms to give the biggest values: δf ±0.04F Therefore a maximum error of±4% in the calculated value of F is possible. Small increments 13/15 Adrian Jannetta

14 Sensitivity Analysis The chain rule for small changes is useful for quantifying the contributions of each variable change to the overall change in the function. Sensitivity analysis Suppose a cylinder with radius r= 10 cm and h= 1 cm is to be constructed by a machine. Is the volume more sensitive to construction errors in the radius or height? The volume of the cylinder is V=πr 2 h so that the chain rule gives: δv 2πrhδr+πr 2 δh Substituting the radius and height gives δv 20πδr + 100πδh This formula tells us that the biggest contribution to volume (for small errors of the same size) comes from the second term; the volume is more sensitive to errors in the heightδh. If the design is changed to r= 1 cm, h=10 cm then we findδv 20πδr+πδh. The volume is now much more sensitive to errors in the radiusδr. Small increments 14/15 Adrian Jannetta

15 Test yourself... Let s finish with some further practice of small increment anatysis. 1 Consider the function P= 4a2 b 5 c 2 Write down an expression forδp in terms of a, b and c. 2 In (1) find the approximate percentage change in P when a and b both decrease by 2% and c decreases by 5%. 3 The current I in a circuit with voltage V and total resistance R is I= V R When V= 12 and R=100, is the current more sensitive to changes in voltage or resistance? Answers: 1 δp 8ab5 δa+ 20a2 b 4 8a2 b 5 δc c 2 c 2 c 3 2 P decreases by 4%. 3 δi 1 12 δv δr. The current is much more sensitive to changes in resistance with these values. Small increments 15/15 Adrian Jannetta

PARTIAL DIFFERENTIATION

PARTIAL DIFFERENTIATION CALCULUS 13 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Partial Differentiation 1/ 15 Adrian Jannetta Functions of many variables The functions and derivatives which